Search Insert Position

📋

Search Insert Position

Edge cases:

</>

IDE

def searchInsert(nums: list[int], target: int) -> int:public int searchInsert(int[] nums, int target)int searchInsert(vector<int> &nums, int target)function searchInsert(nums, target)

def searchInsert(nums, target):
    # Write your solution here
    pass

class Solution {
    public int searchInsert(int[] nums, int target) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int searchInsert(vector<int> &nums, int target) {
    // Write your solution here
    return 0;
}

function searchInsert(nums, target) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: len(nums)Returning length of array unconditionally when target not found, ignoring correct insertion point.✅ Return left boundary index after binary search loop ends instead of len(nums) always.

Wrong: 0Always returning 0 without searching, ignoring target position.✅ Implement binary search to find correct insertion index instead of constant return.

Wrong: Index off by one (e.g., 1 instead of 0 or 3 instead of 4)Off-by-one error in updating left or right pointers or mid calculation in binary search.✅ Use mid = left + (right - left) // 2 and update left = mid + 1 or right = mid correctly.

Wrong: Runtime error or infinite loopIncorrect loop condition or pointer updates causing infinite loop or out-of-bounds access.✅ Ensure while loop condition is left <= right and pointers update properly to avoid infinite loops.

✓

Test Cases

Try testing your code with empty and single element arrays to catch base case bugs.

Check boundary conditions carefully and test targets equal to first and last elements.

Optimize your solution to O(log n) time complexity to handle large inputs efficiently.

t1_01basic

Input{"nums":[1,3,5,6],"target":5}

Expected2

⏱ Performance - must finish in 2000ms

Target 5 is found at index 2.

💡 Think about how binary search narrows down the search space.

💡 Check how to find the lower bound index where target fits or exists.

💡 Return the index where nums[index] >= target or nums.length if none.

Why it failed: Returned wrong index for target found case; fix binary search to return exact match index when nums[mid] == target.

✓ Correctly finds target index using binary search.

t1_02basic

Input{"nums":[1,3,5,6],"target":2}

Expected1

⏱ Performance - must finish in 2000ms

Target 2 not found; should be inserted at index 1 to maintain order.

💡 Consider what happens when target is not in the array.

💡 Binary search should return the insertion point where nums[index] >= target.

💡 Return the left boundary index after binary search completes.

Why it failed: Returned incorrect insertion index for target not found; fix binary search to return left boundary after loop ends.

✓ Correctly returns insertion index for target not found.

t2_01edge

Input{"nums":[],"target":5}

Expected0

⏱ Performance - must finish in 2000ms

Empty array; target should be inserted at index 0.

💡 What if the array is empty? Where should target be inserted?

💡 Check if your code handles zero-length arrays without errors.

💡 Return 0 immediately if nums is empty.

Why it failed: Fails on empty array input; add check for empty nums and return 0.

✓ Handles empty array by returning 0 as insertion index.

t2_02edge

Input{"nums":[10],"target":5}

Expected0

⏱ Performance - must finish in 2000ms

Single element array; target less than element, insert at 0.

💡 Test single element arrays with target less than the element.

💡 Binary search should return 0 if target < nums[0].

💡 Ensure left boundary can be zero when target is smaller than first element.

Why it failed: Incorrect insertion index for target less than single element; fix binary search boundaries to allow left=0.

✓ Correctly inserts target at index 0 when less than single element.

t2_03edge

Input{"nums":[10],"target":15}

Expected1

⏱ Performance - must finish in 2000ms

Single element array; target greater than element, insert at 1.

💡 Test single element arrays with target greater than the element.

💡 Binary search should return nums.length if target > nums[last].

💡 Ensure left boundary can move beyond last index for insertion at end.

Why it failed: Incorrect insertion index for target greater than single element; fix binary search to return nums.length when target is larger than all elements.

✓ Correctly inserts target at index 1 when greater than single element.

t2_04edge

Input{"nums":[1,3,5,6],"target":7}

Expected4

⏱ Performance - must finish in 2000ms

Target greater than all elements; insert at end index 4.

💡 Check insertion when target is larger than all elements.

💡 Binary search should return nums.length if target > nums[last].

💡 Return left boundary after binary search loop ends.

Why it failed: Fails to return nums.length for target greater than all elements; fix binary search to handle this case.

✓ Correctly inserts target at end index when greater than all elements.

t3_01corner

Input{"nums":[1,3,5,6],"target":1}

Expected0

⏱ Performance - must finish in 2000ms

Target equals first element; should return index 0.

💡 Check if binary search correctly returns index when target equals first element.

💡 Verify boundary conditions when target == nums[left].

💡 Return left index immediately if nums[left] == target.

Why it failed: Fails to return index 0 when target equals first element; fix binary search to check equality at boundaries.

✓ Correctly returns index 0 for target equal to first element.

t3_02corner

Input{"nums":[1,3,5,6],"target":6}

Expected3

⏱ Performance - must finish in 2000ms

Target equals last element; should return last index 3.

💡 Check if binary search correctly returns index when target equals last element.

💡 Verify boundary conditions when target == nums[right].

💡 Return right index immediately if nums[right] == target.

Why it failed: Fails to return last index when target equals last element; fix binary search to check equality at boundaries.

✓ Correctly returns last index for target equal to last element.

t3_03corner

Input{"nums":[1,3,5,6],"target":4}

Expected2

⏱ Performance - must finish in 2000ms

Target 4 not found; should be inserted at index 2 between 3 and 5.

💡 Beware of off-by-one errors in binary search mid calculation.

💡 Check that mid is calculated as mid = left + (right - left) // 2.

💡 Ensure loop updates left and right correctly to find insertion point.

Why it failed: Off-by-one error causes wrong insertion index; fix mid calculation and boundary updates in binary search.

✓ Correctly finds insertion index 2 for target 4.

t4_01performance

Input{"nums":[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100000],"target":50000}

Expected49999

⏱ Performance - must finish in 2000ms

n=100000, O(log n) binary search must complete within 2 seconds.

💡 Binary search runs in O(log n) time, suitable for large n.

💡 Avoid linear scans which are O(n) and will time out.

💡 Use iterative binary search to minimize overhead.

Why it failed: TLE due to linear scan or inefficient search; fix to use O(log n) binary search.

✓ Efficient O(log n) binary search confirmed by passing performance test.

Practice

1. You are given an array that was originally sorted in ascending order but then rotated at some unknown pivot. The array may contain duplicates. Your task is to find the minimum element efficiently. Which approach best guarantees an optimal solution for this problem?

easy

A. Use a simple linear scan to find the minimum element by checking all elements.

B. Use a greedy approach that picks the smallest element seen so far while iterating from the start.

C. Use a dynamic programming approach to store minimum values for subarrays and combine results.

D. Use a modified binary search that handles duplicates by shrinking the search space when elements at mid and high are equal.

Solution

1. Step 1: Understand problem constraints

   The array is rotated and may contain duplicates, which complicates the binary search because duplicates can hide the pivot.

2. Step 2: Identify suitable algorithm

   Modified binary search that shrinks the search space when duplicates are encountered ensures correctness and efficiency on average.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Modified binary search handles duplicates correctly [OK]

Hint: Duplicates require shrinking high pointer in binary search [OK]

Common Mistakes:

• Assuming simple binary search works without handling duplicates
• Using greedy or DP unnecessarily

2. You are given an array of positions of balls on a number line and a number m. You want to place m balls in these positions such that the minimum distance between any two balls is maximized. Which algorithmic approach guarantees finding the optimal minimum distance efficiently?

easy

A. Binary search on the answer space combined with a greedy feasibility check

B. Greedy placement by always placing balls at the closest possible positions without backtracking

C. Dynamic programming to consider all subsets of positions and their minimum distances

D. Sorting the positions and then using a sliding window to find the maximum minimum distance

Solution

1. Step 1: Understand the problem structure

   The problem asks to maximize the minimum distance between placed balls, which is a classic optimization problem over a numeric answer space.

2. Step 2: Identify the correct approach

   Binary search on the answer space (possible distances) combined with a greedy feasibility check efficiently finds the maximum minimum distance. Greedy alone or DP are either incorrect or inefficient.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Binary search narrows down distance candidates efficiently [OK]

Hint: Maximize minimum distance -> binary search on answer space [OK]

Common Mistakes:

• Thinking greedy alone suffices without binary search
• Using DP which is inefficient here

3. What is the time complexity of counting occurrences of a target in a sorted array using binary search to find left and right boundaries?

medium

A. O(n) because you must scan the array to count

B. O(log n) because binary search halves the search space each step

C. O(log n * log n) because two binary searches are performed

D. O(1) because built-in functions are optimized

Solution

1. Step 1: Analyze binary search calls

   Two binary searches are performed independently, each O(log n).

2. Step 2: Combine complexities

   Total time is O(log n) + O(log n) = O(log n), since constants are ignored.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Two binary searches do not multiply complexities [OK]

Hint: Two O(log n) calls sum to O(log n) [OK]

Common Mistakes:

• Confusing two calls as O(log n * log n)
• Assuming linear scan is needed
• Believing built-in functions run in O(1)

4. The following code attempts to find the k-th smallest number in an m x n multiplication table. Which line contains a subtle bug that can cause incorrect results for large inputs?

medium

A. Line 4: count = 0

B. Line 6: count += x // i # missing min with n

C. Line 8: if count >= k:

D. Line 13: mid = (low + high) // 2

Solution

1. Step 1: Identify counting logic

   The counting function must cap counts per row by n, since each row has only n elements.

2. Step 2: Analyze line 6

   Line 6 adds x // i without min(n, x // i), causing overcounting when x // i > n, leading to incorrect binary search decisions.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Bug causes wrong counts and incorrect final answer [OK]

Hint: Count per row capped by n to avoid overcounting [OK]

Common Mistakes:

• Forgetting min(n, x//i) in counting
• Incorrect mid calculation

5. Consider the following buggy code snippet for finding the pivot in a rotated sorted array. Which line contains the subtle bug that can cause incorrect pivot detection?

medium

A. if nums[mid] > nums[high]: -- comparison logic is correct

B. mid = (low + high) / 2 -- mid should be integer index, not float

C. low = mid + 1 -- updating low is correct when nums[mid] > nums[high]

D. return low -- returning low as pivot is correct

Solution

1. Step 1: Identify mid calculation issue

   mid is computed using division '/', which returns float in Python 3, causing invalid indexing.

2. Step 2: Impact of float mid

   Using float as index causes runtime error or incorrect behavior; integer division '//' is required.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Changing '/' to '//' fixes the bug [OK]

Hint: mid must be integer index [OK]

Common Mistakes:

• Using '/' instead of '//' in Python
• Misplaced comparison operators